Q30 of 53 Page 59

Calculate the amount of benzoic acid required for preparing 250mL of 0.15M solution in methanol.

Volume of the solution = 250mL = 0.25L

Let the no. Of moles of solute be n


Molarity = No. of moles of solute/volume of solution


⇒ 0.15 = n/0.25


⇒ n = 0.0375moles


Molar mass of C6H5OH = 6×12 + 5×1 + 16 + 1 = 94g


Moles = mass/molar mass


⇒ 0.0375 = m/94


Mass of benzoic acid required = 3.525g.


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